Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 8 - Techniques of Integration - 8.3 Trigonometric Substitution - Exercises - Page 410: 25


$$\frac{1}{16} \sec ^{-1} \frac{z}{2}+\frac{\sqrt{z^{2}-4}}{8 z^{2}}+C$$

Work Step by Step

Given $$\int \frac{d z}{z^{3} \sqrt{z^{2}-4}}$$ Let $$z=2\sec u\ \ \ \ \ \to\ \ \ \ dz= 2\sec u\tan udu $$ Then \begin{align*} \int \frac{d z}{z^{3} \sqrt{z^{2}-4}}&=\int \frac{ 2\sec u\tan udu }{8\sec^3 u \sqrt{4\sec^{2}u-4}}\\ &=\int \frac{\tan u d u}{4 \sec ^{2}u \sqrt{4\left(\sec ^{2}u-1\right)}}\\ &=\int \frac{\tan u d u}{8 \sec ^{2} u \sqrt{\tan ^{2} u}}\\ &=\frac{1}{8} \int \frac{d u}{\sec ^{2} u}\\ &=\frac{1}{8} \int \cos ^{2} u d u\\ &=\frac{1}{16} \int (1+\cos 2 u)du\\ &=\frac{1}{16}\left(u+\frac{\sin 2u}{2}\right)+C\\ &= \frac{1}{16}\left(u+ \sin u\cos u\right)+C\\ &=\frac{1}{16}\left[\sec ^{-1} \frac{z}{2}+\frac{\sqrt{z^{2}-4}}{z} \cdot \frac{2}{z}\right]+C\\ &=\frac{1}{16} \sec ^{-1} \frac{z}{2}+\frac{\sqrt{z^{2}-4}}{8 z^{2}}+C \end{align*}
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